
\prob{00A6}{三角函数正交性}

对于两不同正整数$n, k$，求证：
\begin{align*}
  \int_{-\pi}^\pi \sin nx \dif x = \int_{-\pi}^\pi \cos nx \dif x &= 0 \\
  \int_{-\pi}^\pi \sin nx \cos nx \dif x &= 0 \\
  \int_{-\pi}^\pi \sin nx \cos kx \dif x &= 0 \\
  \int_{-\pi}^\pi \sin nx \sin kx \dif x &= 0 \\
  \int_{-\pi}^\pi \cos nx \cos kx \dif x &= 0
\end{align*}
\problabels{yellow/微积分, green/证明题}

\subsection{积化和差}

首先，显然有
\begin{align*}
  & \int_{-\pi}^\pi \sin nx \dif x = \frac1n \int_{-\pi}^\pi \sin nx \dif nx \\
  ={}& \left.\frac1n \cdot (-\cos nx)\right|_{-\pi}^\pi = \left.-\frac1n\cos nx\right|_{-\pi}^\pi = 0
\end{align*}
同理得
\[ \int_{-\pi}^\pi \cos nx \dif x = 0 \]

由积化和差得
\begin{align*}
  \sin nx \cos kx &= \frac12(\sin(n + k)x + \sin(n - k)x) \\
  \sin nx \sin kx &= -\frac12(\cos(n + k)x - \cos(n - k)x) \\
  \cos nx \cos kx &= \frac12(\cos(n + k)x + \cos(n - k)x) \\
\end{align*}
于是当$n \ne k$时，有
\begin{align*}
  & \int_{-\pi}^\pi \sin nx \cos kx \dif x \\
  ={}& \frac12\int_{-\pi}^\pi (\sin(n + k)x + \sin(n - k)x) \dif x \\
  ={}& \left.\frac12\left(\frac{\sin(n + k)x}{n + k} + \frac{\sin(n - k)x}{n - k}\right)\right|_{-\pi}^\pi \\
  ={}& \frac12(0 + 0) - \frac12(0 + 0) = 0
\end{align*}
当$n = k$时，
\[ \int_{-\pi}^\pi \sin nx \cos kx \dif x = \frac12\int_{-\pi}^\pi \sin 2nx \dif x = 0 \]
同理得
\begin{align*}
  \int_{-\pi}^\pi \sin nx \sin kx \dif x &= 0 \\
  \int_{-\pi}^\pi \cos nx \cos kx \dif x &= 0
\end{align*}
证毕。
